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This Lesson (Probabilistic analysis of a court verdicts) was created by by ikleyn(52750)
  About ikleyn: Probabilistic analysis of a court verdictsProblem 1A court consists of 3 judges.Two of them, independently of each other, make correct decisions with a probability of 0.5. The third judge agrees with the first two decisions when they match. In case of different opinions of the first two judges, the third judge decides on his own and makes a mistake with a probability of 0.43. What is the probability that the court will not make a mistake if the verdict is decided by a majority vote? Solution
We do not know what cases the judges do consider and what judgment/decisions do they make.
We only know that the decisions can be Right or Wrong.
Therefore, based on given information, we only can make a table of probabilities
for all possible situations.
So, I made this table: it is below.
1 2 3 verdict decided Include (+) Individual
by a majority not include (-) probabilities
for each possible court decision
--------------------------------------------------------------------------------------------
1 R R R ---> R + 0.5*0.5*1
2 W W W ---> W -
3 R W R ---> R + 0.5*0.5*0.57
4 R W W ---> W -
5 W R R ---> R + 0.5*0.5*0.57
6 W R W ---> W -
The leftmost column is for the numbers of lines.
The digits 1, 2 and 3 in the horizontal upper line represent the judges.
"R" represents right decision; "W" represents wrong decision.
The symbols in the table below "1", "2" and "3" symbolize the decisions (R for right, W for wrong).
The symbols in the column named "verdict decided by a majority"
represent the logical consequence of the decisions made in columns 1, 2, and 3.
It is how the court makes its final decision, based on individual decisions of the judges.
The arrows ( ---> ) show the logical implications ("verdict decided by a majority").
Notice that in the table I listed ALL LOGICALLY POSSIBLE situations.
There are NO other possible situations that would be consistent with the problem.
According to the problem, the question is about the probability
of the final court's decision to be right. So, in column "include or not include"
I write "+" for right decisions to include them into the final count
or "-" for wrong decisions to NOT include them into the final count.
In the rightmost column, I calculated the probabilities for each
possible RIGHT decision of the curt.
From this consideration, the final probability of the right verdict of the court is
0.5*0.5*0.57 +0.5*0.5*0.57 = 0.535. ANSWER
This my solution is written as a description of an ALGORITHM calculating the desired probability based on given input data. Problem 2Three judges with a majority vote have to make a decision.The judges decide independently of each other. Based on previous experience, it is known that the probabilities of each judge making the correct decision are 0.83, 0.83, 0.87. (a) What is the probability that the final decision will be correct? (b) If the final decision was correct, what is the probability that the first judge made the right decision? Solution
We do not know what cases the judges do consider and what judgment/decisions do they make.
We only know that the decisions can be Right or Wrong.
Therefore, based on given information, we only can make a table of probabilities
for all possible situations.
So, I made this table: it is below.
1 2 3 verdict decided Include (+) Individual
by a majority not include (-) probabilities
for each possible court decision
--------------------------------------------------------------------------------------------
1 R R R ---> R + 0.83*0.83*0.87
2 R R W ---> R + 0.83*0.83*(1-0.87)
3 W W R ---> W -
4 W W W ---> W -
5 R W R ---> R + 0.83*(1-0.83)*0.87
6 R W W ---> W -
7 W R R ---> R + (1-0.83)*0.83*0.87
8 W R W ---> W -
The leftmost column is for the numbers of lines.
The digits 1, 2 and 3 in the horizontal upper line represent the judges.
"R" represents right decision; "W" represents wrong decision.
The symbols in the table below "1", "2" and "3" symbolize the decisions (R for right, W for wrong).
The symbols in the column named "verdict decided by a majority"
represent the logical consequence of the decisions made in columns 1, 2, and 3.
It is how the court makes its final decision, based on individual decisions of the judges.
The arrows ( ---> ) show the logical implications ("verdict decided by a majority").
Notice that in the table I listed ALL LOGICALLY POSSIBLE situations.
There are NO other possible situations that would be consistent with the problem.
According to the problem, the question is about the probability
of the final court's decision to be right. So, in column "include or not include"
I write "+" for right decisions to include them into the final count
or "-" for wrong decisions to NOT include them into the final count.
In the rightmost column, I calculated the probabilities for each
possible RIGHT decision of the curt.
From this consideration, the final probability of the right verdict of the court is
0.83*0.83*0.87 + 0.83*0.83*(1-0.87) + 0.83*(1-0.83)*0.87 + (1-0.83)*0.83*0.87 = 0.934414 (precise value). (*) ANSWER
You may round it in any way you want.
Thus part (a) is complete. The post-solution note: Beeing familiar with this detailed explanation and its logic, and looking in the structure of the formula (*), you may produce much shorter solution in your head on your own in couple of lines. ------------------- This my solution is written as a description of an ALGORITHM calculating the desired probability based on given input data. ///////////////////////
Now for part (b), it asks about the conditional probability.
It is the ratio of two values.
The denominator is the value of the part (a) above.
The numerator is that three addends of the formula (*)
0.83*0.83*0.87 + 0.83*0.83*(1-0.87) + 0.83*(1-0.83)*0.87,
where the first judge was right.
So, the answer for (b) is
At this point, the problem is solved in full. My other lessons on Probability in this section are - A company bids on two separate contracts - Advanced probability problems related to combinations - Using cumulative sums and relevant standard functions to solve problems on Binomial Distributions - Miscellaneous binomial distribution problems - Binomial distribution problems on overbooking flights - Advanced probability problems on binomial distribution - Accepting/rejecting shipments via acceptance procedures - Probabilistic analysis of testing procedures in health care - Probabilistic analysis of testing procedures in industry - Probability problems on games - Probability problem on winning a many-rounds game - A Math circle level probability problem on a lottery game - Upper league entertainment probability problems - Upper league probability problems on Stars and Bars methodology - Upper league problem to maximize winning in a game with 20-sided rolling die - Upper league problems on conditional probability - Upper League geometric probability problems - Probability problems on long chains of related events - Probability problems similar to a coinciding birthdays problem - Using empirical rules to determine normal distribution probabilities - OVERVIEW of lessons on Probability, section 3 Use this file/link OVERVIEW of lessons on Probability to navigate over all my lessons on Probability problems (section 1) in this site. Use this file/link OVERVIEW of my additional lessons on Probability to navigate over all my lessons on additional Probability problems (section 2) in this site. Use this file/link ALGEBRA-II - YOUR ONLINE TEXTBOOK to navigate over all topics and lessons of the online textbook ALGEBRA-II. This lesson has been accessed 929 times. |
